Arrays ELEC 206 Computer Applications for Electrical Engineers Dr. Ron Hayne

Arrays

ELEC 206

Computer Applications for Electrical Engineers

Dr. Ron Hayne

206_C6 2

Outline

Arrays Statistical Measurements Functions Revisited

206_C6 3

Arrays

One-dimensional array List of values Arranged in either a row or a column

Offsets (Subscripts) distinguish between elements in the array Identifier holds address of first element Offsets always start at 0

206_C6 4

Definition and Initialization

Declaration double x[4];

Initialization double x[4] = {1.2, -2.4, 0.8, 6.1}; int t[100] = {0}; char v[] = {'a', 'e', 'i', 'o', 'u'};

Loops double g[21];

for (int k=0; k<=20; k++)

{

g[k] = k*0.5;

}

206_C6 5

Data Files

double time[10], motion[10];

ifstream sensor3("sensor3.dat");

...

if(!sensor3.fail())

{

for (int k=0; k<=9; k++)

{

sensor3 >> time[k] >> motion[k];

}

}

206_C6 6

Function Arguments

Passing array information to a function Two parameters usually used

Specific arrayNumber of elements in the array

Always call by referenceAddress of the array

/* This program reads values from a data file and *//* calls a function to determine the maximum value *//* with a function. */

#include <iostream>#include <fstream>#include <string>using namespace std;

// Define constants and declare function prototypes.const int N=100;double maxval(double x[], int n);

int main(){ // Declare objects. int npts=0; double y[N]; string filename; ifstream lab;

...

// Read a data value from the file. while (npts <= (N-1) && lab >> y[npts] ) { npts++; // Increment npts. }

// Find and print the maximum value. cout << "Maximum value: " << maxval(y,npts) << endl;

// Close file and exit program. lab.close();

}return 0;}

/* This function returns the maximum *//* value in the array x with n elements. */

double maxval(double x[], int n){ // Declare local objects. double max_x;

// Determine maximum value in the array. max_x = x[0]; for (int k=1; k<=n-1; k++) { if (x[k] > max_x) max_x = x[k]; }

// Return maximum value. / return max_x;}

206_C6 10

Statistical Measurements

Maximum Minimum Mean (average) Median (middle) Variance

Average squared deviation from the mean

Standard Deviation Square root of the variance

206_C6 11

Mean

double mean(double x[], int n)

{

double sum(0);

for (int k=0; k<=n-1; k++)

{

sum += x[k];

}

return sum/n;

}

206_C6 12

Variance

double variance(double x[], int n)

{

double sum(0), mu;

mu = mean(x,n);

for (int k=0; k<=n-1; k++)

{

sum += (x[k] - mu)*(x[k] - mu);

}

return sum/(n-1);

}

206_C6 13

Function Overloading

Function name can have more than one definition Each function definition has a unique function signature Each function call looks for a function prototype with a

matching function signature

double maxval(double x[], int n); int maxval(int x[], int n); char maxval(char x[], int n);

206_C6 14

Function Templates

Generic algorithm for a function that is not tied to a specific data type

template <class Data_type>

Data_type minval(Data_type x[], int n)

{

Data_type minx;

...

206_C6 15

Summary

Arrays Statistical Measurements Functions Revisited

206_C6 16

Problem Solving Applied

Speech Signal Analysis Problem Statement

Compute the following stastical measurements for a speech utterance: average power, average magnitude, and number of zero crossings.

Input/Output Description

Zero.dat

Average power

Average magnitude

Zero crossings

206_C6 17

Problem Solving Applied

Hand Exampletest.dat

2.5 8.2 -1.1 -0.2 1.5Average power = 15.398Average magnitude = 2.7Number of zero crossings = 2

206_C6 18

Problem Solving Applied

Algorithm Developmentmain

read speech signal from data file and determine number of points, n

compute statistical measurements using functions

ave_power(x, n) set sum to zero for k: add x[k]2 to sum return sum/n

206_C6 19

Problem Solving Applied

Algorithm Developmentave_mag(x, n)

set sum to zero for k: add |x[k]| to sum return sum/n

crossings(x, n) set count to zero for k: if x[k] * x[k+1] < 0, increment count return count

/*---------------------------------------------------*//* This program computes a set of statistical */ /* measurements from a speech signal. */

#include <cstdlib>#include <iostream>#include <fstream>#include <string> #include <cmath> using namespace std;

// Declare function prototypes and define constants.double ave_power(double x[], int n);double ave_magn(double x[], int n);int crossings(double x[], int n);

const int MAXIMUM = 2500;

int main(){ // Declare objects. int npts=0; double speech[MAXIMUM]; string filename; ifstream file_1;

// Prompt user for file name and open file. cout << "Enter filename "; cin >> filename; file_1.open(filename.c_str()); if( file_1.fail() ) { cout << "error opening file " << filename << endl; return 1; }

// Read information from a data file. while (npts <= MAXIMUM-1 && file_1 >> speech[npts]) { npts++; }

// Compute and print statistics. cout << "Statistics \n"; cout << "\taverage power: " << ave_power(speech,npts) << endl; cout << "\taverage magnitude: " << ave_magn(speech,npts) << endl; cout << "\tzero crossings: " << crossings(speech,npts) << endl; // Close file and exit program. file_1.close(); system("PAUSE"); return 0;}

/*---------------------------------------------------*//* This function returns the average power *//* of an array x with n elements. */

double ave_power(double x[], int n){ // Declare and initialize objects. double sum=0;

// Determine average power. for (int k=0; k<=n-1; k++) { sum += x[k]*x[k]; }

// Return average power. return sum/n;}

/*---------------------------------------------------*//* This function returns the average magnitude *//* of an array x with n values. */

double ave_magn(double x[], int n){ // Declare and initialize objects. double sum=0;

// Determine average magnitude. for (int k=0; k<=n-1; k++) { sum += abs(x[k]); }

// Return average magnitude. return sum/n;}

/*---------------------------------------------------*//* This function returns a count of the number *//* of zero crossings in an array x with n values. */

int crossings(double x[], int n){ // Declare and initialize objects. int count=0;

// Determine number of zero crossings. for (int k=0; k<=n-2; k++) { if (x[k]*x[k+1] < 0) count++; }

// Return number of zero crossings. return count;}/*---------------------------------------------------*/

206_C6 26

Testing

206_C6 27

Testing

206_C6 28

Sorting Algorithms

Sorting Arranging in ascending (descending) order Not one "best" algorithm

Depends on characteristics of data Selection Sort

Find smallest value from current position to endSwap smallest with current positionMove to next position

#include <iostream>#include <cstdlib>using namespace std;

void sort(double x[], int n);void display(double x[], int n);const int N = 6;

int main(){ double data[N] = {5.0, 3.0, 12.0, 8.0, 1.0, 9.0}; cout << "Initial data" << endl; display(data, N); sort(data, N); cout << "Sorted data" << endl; display(data, N); ...

void sort(double x[], int n) // Selection sort{ int m; // marker double hold; for (int k=0; k<=n-2; k++) { m = k; // Find smallest value for (int j=k+1; j<=n-1; j++) { if (x[j] < x[m]) m = j; } hold = x[m]; // Swap smallest with current x[m] = x[k]; x[k] = hold; cout << "After k=" << k << endl; display(x, n); }}

206_C6 31

Testing

206_C6 32

Summary

Arrays Statistical Measurements Functions Revisited Problem Solving Applied Selection Sort End of Chapter Summary

C++ Statements Style Notes Debugging Notes